The document provides data from a test on a flat-walled furnace with inner and outer brick walls of unknown conductivity. The temperatures at various points within the furnace walls are given. Using this data and the thickness of each material, the conductivity of each brick type is calculated. The percentage of heat loss saved by adding 5cm of magnesia insulation to the furnace is calculated to be 52.73%.

1. CONDUCTION Problem Set Alejandro, Jan Rannel Granada, Gat Hendrix, Jamie Salcedo, Angel Frances Villaflor, Kaye

2. STEADY STATE APPLICATIONS 5. The following data was obtained in a test on a flat-walled furnace, the linings of which consist of a 11.5 cm. non-corrosive brick of unknown conductivity and the outer wall of 20 cm clay brick, also of unknown conductivity. The temperature of the inner wall (flame side) was found to be 735°C and that of the outer wall 185°C. This furnace was lagged with 5 cm of magnesia (k=0.07 W/mK) thermocouples inserted at various points and the ff. data taken: Calculate the percentage of heat loss that is saved by the lagging. 88°C Temp of the outer surface of magnesia 475°C Temp at the junction of ordinary brick and magnesia 700°C Temp at the junction of brick layers 735°C Temp of inner wall (flame side)

3. STEADY STATE APPLICATIONS Magnesia

4. Solution: Basis: Am = 1m2 From the data of composition with magnesia: q = ∆T/R R = ∆x/(km)(Am) q = ∆T(km)(Am)/ ∆x Series: q1 = q2 = q3 =qtotal q of magnesia(q3) = (475-88)K(0.07W/mK)( 1m2) = 541.8 W 0.05m q of magnesia = q of non-corrosive brick = q of clay brick = q total = 541.2 W km = q(∆x)/(∆T)(Am) km of non-corrosive brick = (541.8W)(0.115m)/(735-700)( 1m2) = 1.7802 W/mK km of clay brick = (541.8 W)(0.20m)/(700-475) ( 1m2) = 0.4816 W/mK non-corrosive brick clay brick Magnesia 735°C 700°C 475°C 88°C

5. Using the data in the composition without the magnesia: Series: Rtotal = R1 + R2 + R3… Getting the R: R = ∆x/(km)(Am) R non-corrosive brick = (0.115m) /(1.7802 W/mK) ( 1m2) = 0.06460 K/W R of clay brick = (0.20m)/(0.4816 W/mK)( 1m2) = 0.4153 K/W Rtotal = 0.4799K/W q total = (735-185)K/(0.4799 K/W) = 1146.07 W percentage of heat loss that is saved by the lagging: (q total without the magnesia – q total with magnesia) x 100% q total without magnesia (1146.07W-541,8W) x 100% = 52.73% 1146.07W non-corrosive brick clay brick 735°C 185°C

6. UNSTEADY STATE APPLICATIONS 13. (US)What length of time is required to heat a 2.5 cm diameter, 5 cm long copper cylinder ( k = 380 W/mK ) from 16 oC to 93 oC in a furnace whose temperature is 315 oC? The value of h at the cylinder surface is 204 KJ/hm2K and its thermal diffusivity is 0.186 m2/s.

7. UNSTEADY STATE APPLICATIONS 13. (US)What length of time is required to heat a 2.5 cm diameter, 5 cm long copper cylinder ( k = 380 W/mK ) from 16 oC to 93 oC in a furnace whose temperature is 315 oC? The value of h at the cylinder surface is 204 KJ/hm2K and its thermal diffusivity is 0.186 m2/s. Given: X = 2.5cm To = 16 oC h = 204 kJ/hm2K L = 5cm T = 93 oC α = 0.186 m2/s k = 38.0 W/mK T1 = 315 oC

8. UNSTEADY STATE APPLICATIONS Solution: h = 204 _kJ_ x 1000J x 1h___ = 56.67 W/m2K hm2K 1 kJ 3600 s X1 = 2.5/100 = 0.0125m 2 Y = 315 – 93 = 0.74 315 – 16 m = k__ = 38.0 W/mK__________ = 53.64 hX1 56.67 W/m2K (0.0125m) From the Heisler Chart(Cylinder): X = 10.1 = αt__ = (0.186) t t = 0.0085s X12 (0.0125)2

10. Geankoplis Problem 4.2-3. Removal of Heat from a Bath . A cooling coil of 1.0ft of 308 stainless steel have an average thermal conductivity of 15.23 W/mK, inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40oF and is 80oF on the outside. Calculate the heat removal in BTU/s and watts. Where k is in BTU/ h ft oF and T is in oF.

11. 4.2-3. Removal of Heat from a Bath . A cooling coil of 1.0ft of 308 stainless steel have an average thermal conductivity of 15.23 W/mK, inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40oF and is 80oF on the outside. Calculate the heat removal in BTU/s and watts. Where k is in BTU/ h ft oF and T is in oF. Given : L = 10 ft = 0.3048m k = 15.23 W/mK Di = 0.25 in. = 6.35x10-3m Do = 0.40 in. = 0.01016m T1­ = 40oF = 4.44oC + 273K = 277.44K T2 = 80oF = 26.67oC + 273K = 299.67K

12. 4.2-3. Removal of Heat from a Bath . A cooling coil of 1.0ft of 308 stainless steel have an average thermal conductivity of 15.23 W/mK, inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40oF and is 80oF on the outside. Calculate the heat removal in BTU/s and watts. Where k is in BTU/ h ft oF and T is in oF. Solution: X = r2 - r1 = (0.01016/2) – (6.35x10-3/2) =1.905x10-3m Dlm = 0.01016-6.35 x10-3­___ = 8.11x10-3 ln (0.1016/6.35x10-3) Am = πlDlm q = T1 - T2 = (299.67 – 277.44)K______ X__ 1.905x10-3m________________ = 1380.16watts or J/s kmAm 15.23W/mK[(π)(0.3048m)(8.11x10-3)]m2 q = 1380.16watts or J x 1BTU = 1.31 BTU/s s 1055J

13. Geankoplis Problem 5.3-5. Cooling a Slab of Meat. A slab of meat 25.4 mm thick originally at a uniform temperature of 10cC is to be cooked from both sides until the center reaches 121cC in an oven at 177cC. The convection coefficient can be assumed constant at 25.6 W/m2-K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/m-K and the thermal diffusivity 5.85x10-4m2/h. Use the Heisler Chart.

14. 5.3-5. Cooling a Slab of Meat. A slab of meat 25.4 mm thick originally at a uniform temperature of 10cC is to be cooked from both sides until the center reaches 121cC in an oven at 177cC. The convection coefficient can be assumed constant at 25.6 W/m2-K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/m-K and the thermal diffusivity 5.85x10-4m2/h. Use the Heisler Chart. Given: X = 25.4mm = 0.0254m To = 10oC T = 121 cC (X = 0,center) T1 = 177 cC h = 25.6 W/m2-K k = 0.69 W/m-K α = 5.85x10-4m2/h

15. 5.3-5. Cooling a Slab of Meat. A slab of meat 25.4 mm thick originally at a uniform temperature of 10cC is to be cooked from both sides until the center reaches 121cC in an oven at 177cC. The convection coefficient can be assumed constant at 25.6 W/m2-K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/m-K and the thermal diffusivity 5.85x10-4m2/h. Use the Heisler Chart. Solution: X1 = 0.0254/2 = 0.0127m Y = T1 – T = 177 – 121 = 0.34 T1 - To 177 – 10 m = k__ = 0.69.0 W/mK__________ = 2.1 hX1 (25.6 W/m2K)(0..0127m) n = 0___ = 0 0.0127 From the Heisler Chart (Plate): X = 3.2 = αt__ = ( 5.85x10-4m2/h ) t t = 0.88h x 3600s = 3168s X12 ( 0.0127 )2 1h